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**Fields & Effects**

**Capacitors 2**

energy stored |

__Energy Stored in a Capacitor__

The energy stored in a capacitor is in the form of electrical potential energy. This is has two components:

work done in negative plate
work done in positive plate |

Consider a * partially* charged capacitor, with a p.d. of

*volts across it and charge*

**V***on it plates.*

**Q**

Now during charging, consider a small charge * δQ *moving from one plate to another.

If * δQ* is very small then the increase in p.d. is also very small. So

*may be considered approximately the same.*

**V**

Hence the work done * δW *, is given by :

Recalling the equation for capacitance * C *and rearranging to make

*the subject :*

**V**

Substituting for * V *in our original equation.

The total work done is the area under the curve for 0 to * Q_{o}*, where

*is the maximum charge stored.*

**Q**_{o}

Integrating between the limits gives the result :

In the general case, we can write * Q* =

*.*

**Q**_{o}

Recalling the capacitor equation again and making * Q* the subject :

Substituting for * Q* into the equation for

*:*

**W**

Now substituting for * CV = VC = Q *,

Summarizing,

__Capacitors in Parallel__

Capacitors in parallel have the same p.d. across them.

Writing * Q = CV *for each capacitor and adding :

Putting * Q_{T}* as the total charge,

Hence,

Factorizing,

But

Therefore,

__Capacitors in Series__

The battery removes charge **Q ^{-}** from plate '

**a**' and deposits it on plate '

**f**'.

Plate 'a' is therefore left with a charge **Q ^{+}**on its plates.

**Each charged plate then induces an opposite charge in its opposing plate.**

The central capacitor **C _{2}** has a positive charge on plate '

**c**' because electrons are removed from it to make plate '

**b**' negative. Plate '

**d**' is made negative by induction with '

**c**'.

Making the p.d. * V* the subject for each capacitor and adding:

Summing the p.d. around the circuit, the supply p.d. is * V *is given by:

Hence

Recalling that,

By similarity, it follows that :

__Charging & Discharging a Capacitor through a Resistor __

When the switch is closed charge from the capacitor flows through the resistor. The resistance **R** has the effect of limiting this flow.

For a fully charged capacitor of capacitance **C** let the p.d. be **V _{o}** and the charge

**Q**.

_{o}

Consider the p.d. around the circuit at a time * t* seconds from the start of the discharge.

Since there is no net p.d. in the circuit, by **Kirchoff's 2nd law** (relating to p.d. in a circuit) :

Quoting the capacitor equation, with* V* the subject,

Substituting for * V* in our initial equation,

(i

The current * I* is defined as the rate of charge of charge with time,

Remember the **Ohm's Law** equation,

Now, substituting for * V* and

*into equation (i ,*

**I**

Separating the differential operators * dQ* and

**dt**,

Since * Q* =

*when*

**Q**_{o}*= 0 and*

**t***=*

**Q***when*

**Q***=*

**t***, integrating between these limits :*

**t**

Hence,

Therefore,

and,

Changing into exponential form :

Rearranging into a more familiar form,

Substituting for * Q* =

*and*

**VC***=*

**Q**_{o}*, cancelling the*

**V**_{o}C*'s*

**C**

So both the charge and the p.d. decrease at an exponetial rate.

When the elapsed time * t *=

*the*

**CR****charge remaining is approx. 37% of the original amount**.

The p.d. across the resistor * V_{R}* and the current

*through it are given by:*

**I**

**NB** The significance of the **minus signs** in these equations.

This is a consequence of **Kirchoff's 2nd law**. Remember how the p.d. across the resistor and the capacitor are related.

If * V_{R}* is taken to the other side of the equation it becomes negative. Hence one p.d. is the negative of the other.

The second equation (* I*) is obtained from the first by substituting

*and rearranging.*

**V**_{R}= IR

The curve of * Q* vs

*for charging is :*

**t**

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